(g) A thin, 1.44-kg slab of copper at 18.0°C is set sliding at 2.90 m/s over an identical stationary slab at the same temperature. Friction quickly stops the motion. Assuming no energy is transferred to the environment by heat, find the change in temperature of both objects. (Assume the specific heat of copper is 387 J/kg · °C.) ΔΤ sliding slab The correct answer is not zero.°C ΔΤ. stationary slab The correct answer is not zero.°C (h) Evaluate Q and AEnt for the sliding slab and AE, mech for the two-slab system. %3D ΔΕ int ΔΕ mech
(g) A thin, 1.44-kg slab of copper at 18.0°C is set sliding at 2.90 m/s over an identical stationary slab at the same temperature. Friction quickly stops the motion. Assuming no energy is transferred to the environment by heat, find the change in temperature of both objects. (Assume the specific heat of copper is 387 J/kg · °C.) ΔΤ sliding slab The correct answer is not zero.°C ΔΤ. stationary slab The correct answer is not zero.°C (h) Evaluate Q and AEnt for the sliding slab and AE, mech for the two-slab system. %3D ΔΕ int ΔΕ mech
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Transcribed Image Text:(g) A thin, 1.44-kg slab of copper at 18.0°C is set sliding at 2.90 m/s over an identical stationary slab at the same temperature.
Friction quickly stops the motion. Assuming no energy is transferred to the environment by heat, find the change in temperature
of both objects. (Assume the specific heat of copper is 387 J/kg · °C.)
ΔΤ.
sliding slab
The correct answer is not zero.°C
ΔΤ.
stationary slab
The correct answer is not zero.°C
(h) Evaluate Q and AE,
int
for the sliding slab and AE,
'mech
for the two-slab system.
ΔΕ ,
int
ΔΕ.
mech
(i) Evaluate Q and AE,
int
for the stationary slab.
ΔΕ.
int
Expert Solution
Step 1
(g)
Given:
The mass of the slab of copper is 1.44 kg.
The temperature is .
The speed of the slab is .
The specific heat of copper is .
Introduction:
Heat is the form of energy transferred between two materials that are at different temperatures. Heat always flows from the material with higher temperature to the material with lower temperature until thermal equilibrium is reached.
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